# Maths Assignment

Topics: Differential equation, Mathematics, Terminal velocity Pages: 9 (1521 words) Published: August 30, 2013
Item 4B
Item 4B

Rachel Reiser
Maths C

Rachel Reiser
Maths C

Question 1
ab1+f'(x)2 dx

y = acosh(xa)

If:
coshx=12ex+e-x
Then:
cosh(xa) = 12(exa+e-xa)

y = acosh(xa)

y=a(exa+e-xa)2
y=a(exa+e-xa)2
dydx=f'x=ddxa(exa+e-xa)2

dydx=f'x=ddx12aexa+e-xa
f'x=12a1aexa+-1ae-xa
f'x=exa-e-xa2

f'x2=exa-e-xa22
f'x2=(12exa-12e-xa)(12exa-12e-xa)
f'x2=14e2xa-14e0-14e0+14e-2xa
f'x2=14e2xa-12+14e-2xa
f'x2=14e2xa-2+e-2xa
Assuming the catenary is symmetrical, the entire length of the wire is equal to two multiplied by the length between x = b and x = a (or x = 0) Entire arc length =2×0b1+f'(x)2 dx
Substitute the value for f'x2 into the formula above.
L =2×0b1+14e2xa-2+e-2xa dx
L =2×0b44+14e2xa-2+e-2xa dx
L =2×0b144+e2xa-2+e-2xa dx
L =2×0b142+e2xa+e-2xa dx
L=2×0b142+e2xa+e-2xa dx
L=2×140b2+e2xa+e-2xa dx
L=2×120b2+e2xa+e-2xa dx
Note: rule of perfect squares 2+e2xa+e-2xa=(exa+e-xa)2
L=0b(exa+e-xa)2 dx
L=0bexa+e-xa dx
L=aexa-ae-xa0b
L=aeba-ae-ba-(ae0a-ae-0a)
L=aeba-ae-ba-(a-a)
L=aeba-ae-ba
L=a(eba-e-ba)
∴ The length of the arc in terms of a and b is L=aeba-e-ba

Question 2
mdvdt=mg-kv2
Where:
m=mass of body
v=velocity
g=acceleration due to gravity (9.8m/s)
k=constant related to size and shape of object
mdvdt=mg-kv2
mdv=mg-kv2 ×dt
mdvmg-kv2=dt
dt=mmg-kv2dv
dt=m1mg-kv2dv

mg-kv2 may also be expressed as mg2-kv22
=mg2-kv22
=mg2-kv2
∴ Difference of two squares:
mg+kvmg-kv

∴dt=m1mg+kvmg-kvdv
RHS=m1mg+kvmg-kvdv
let 1mg+kvmg-kv=Amg+vk+Bmg-vk
∴1+0v=mg-vkA+mg+vkB
1+0v=mgA-vkA+mgB+vkB
1+0v=mgA+mgB+vkB-kA

∴1=mgA+mgB0=kB-kA
1=mg(A+B)0=k(B-A)
A+B=1mg --- 1 0=B-A --- 2
A=B --- 2
Sub 2 into 1:
2A=1mg
A=12mg
∴A=12mg B=12mg
1mg+kvmg-kv=Amg+vk+Bmg-vk
∴1mg+kvmg-kv=12mgmg+vk+12mgmg-vk
1mg+kvmg-kv=12mgmg+vk+12mgmg-vk
1mg+kvmg-kvdv=12mgmg+vkdv+12mgmg-vkdv

dt=m1mg+kvmg-kvdv

∴dt=m12mgmg+vk+12mgmg-vkdv
dt=m12mg+2vmgk+12mg-2vmgkdv
dt=m21mg+vmgk+1mg-vmgkdv
t+c=m21mgklnmg+vmgk-1mgklnmg-vmgk
t+c=m21mgklnmg+vmgk-lnmg-vmgk
t+c=m21mgklnmg+vmgkmg-vmgk
t+c=m2mgklnmg+vmgkmg-vmgk
Find c by subbing in v=0, t=0
0+c=m2mgklnmg+0mgkmg-0mgk
c=m2mgklnmgmg
c=m2mgkln1
c=m2mgk×0
∴c=0
∴t=m2mgklnmg+vmgkmg-vmgk
2tmgkm=lnmg+vmgkmg-vmgk
2tmgkm2=lnmg+vmgkmg-vmgk
2tgkm=lnmg+vmgkmg-vmgk
e2tgkm=mg+vmgkmg-vmgk
(mg-vmgk)e2tgkm=mg+vmgk
mge2tgkm-vmgke2tgkm=mg+vmgk
vmgk+vmgke2tgkm=mge2tgkm-mg
vmgk+mgke2tgkm=mge2tgkm-1
v=mge2tgkm-1mgk+mgke2tgkm
v=mge2tgkm-1mgk1+e2tgkm
v=mg2e2tgkm-1mgk1+e2tgkm
v=mge2tgkm-1k1+e2tgkm
v=mgke2tgkm-1e2tgkm+1

v=limt→∞mgke2tgkm-1e2tgkm+1
v=limt→∞mgke2tgkm-1e2tgkm+1÷e2tgkme2tgkm
v=limt→∞mgk1-e-2tgkm1+e-2tgkm
v=limt→∞mgk1-e-∞1+e-∞
e-∞→0
v=mgk1-01+0
v=mgk×1
v=mgk
Therefore the limiting of the body is mgk.

Terminal velocity (i.e. acceleration = 0 m/s2)
Acceleration =dvdt
Where:
v=velocity

mdvdt=mg-kv2
When acceleration =o:
m×0=mg-kv2
0=mg-kv2
kv2=mg
v=mgk

* Let terminal velocity (v)= 12m/s2
Body mass (m) = 30kg
Gravity (g) = 9.8 m/s2

v=mgk
12=30×9.8k
122=294k
k=294144
k=2.04167

v=mgke2tgkm-1e2tgkm+1

Sub values for k, m and g into equation:
v=30×9.82.04167e2t9.8×2.0416730-1e2t9.8×2.0416730+1

Graphical analysis:

Effect of ‘k’ with respect to proportion principles
Terminal velocity:
v=mgk
Where the principle of proportions states that:
y∝x
∴y=kx
Where k is a constant.
v=mgk
In this case mg is the constant (i.e. has a known numerical value) Where y=v, x=k
y=kx
Translates to:
y=1xmg
Where y=y, k=mg, x=1k
∴as y∝1x
v=1k
Therefore terminal velocity is inversely proportional to k.

Question 3
t=0, P=12000
t=200, P=200
dPdt=110P1-P200001-αft
Where:
α=constant which takes into account the effect of introduced predators ft=function of time

General solution to differential equation given above for the case where f(t) = t: dP=110P1-P200001-αft dt
dP=P10-P22000001-αft dt
dPP10-P2200000=1-αft dt
Sub in f(t)...